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A mass attached to two horizontal springs, vertical motion

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roman15
#1
Mar6-11, 09:41 PM
P: 70
1. The problem statement, all variables and given/known data
A mass is connected between two identical springs with the same spring constant, k. The equilibrium length of each spring is L, but they are stretched to 2L when the mass is in equilibrium. Ignoring gravity, find the equations of motion and then angular frequency of oscillation, when the mass is displaced by a small distance +/-x.
| |
|______m______|
| |
this is what the system looks like, with each string 2L in the picture. The x direction is vertical


2. Relevant equations
F=-kx
a(x)=-(w^2)(x)

3. The attempt at a solution
from examining the forces I got that a(x)=(-2kxsin(theta))/m
i didnt know what to do next
would w=squareroot(2k/m)
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Andrew Mason
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Mar7-11, 05:03 AM
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Quote Quote by roman15 View Post
1. The problem statement, all variables and given/known data
A mass is connected between two identical springs with the same spring constant, k. The equilibrium length of each spring is L, but they are stretched to 2L when the mass is in equilibrium. Ignoring gravity, find the equations of motion and then angular frequency of oscillation, when the mass is displaced by a small distance +/-x.
| |
|______m______|
| |
this is what the system looks like, with each string 2L in the picture. The x direction is vertical
I am having difficulty understanding the problem. Are the two springs in series or parallel? Are they both above the mass?

AM
roman15
#3
Mar7-11, 06:20 PM
P: 70
here is an attached picture of what the system looks like, sorry for the confusion
the only difference from this picture is that the mass moves up and down, and the vertical direction is the x axis
Attached Thumbnails
physics pic.gif  

Andrew Mason
#4
Mar7-11, 08:43 PM
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A mass attached to two horizontal springs, vertical motion

Quote Quote by roman15 View Post
here is an attached picture of what the system looks like, sorry for the confusion
the only difference from this picture is that the mass moves up and down, and the vertical direction is the x axis
I am still puzzled. I don't see how you can "ignore gravity".

Do a freebody diagram for m in the supposed equilibrium position. What are the downward forces on m? What is the upward force? How can this be equilibrium?

If m is on a horizontal surface, you can stretch the springs by an equal amount and achieve equilibrium. But when you turn it vertically, the springs will not be stretched equally.

AM
roman15
#5
Mar7-11, 09:51 PM
P: 70
i dont know, but were asked to just ignore gravity so the only forces acting on it when you displace the mass are the two forces of the springs, and no there isnt a horizontal surface
Andrew Mason
#6
Mar8-11, 06:45 AM
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Quote Quote by roman15 View Post
i dont know, but were asked to just ignore gravity so the only forces acting on it when you displace the mass are the two forces of the springs, and no there isnt a horizontal surface
Ok. We will assume it is in orbit in space
from examining the forces I got that a(x)=(-2kxsin(theta))/m
i didnt know what to do next
would w=squareroot(2k/m)
Yes.

Your equation is a differential equation that has a general solution of:

[tex]x = A\sin{(\omega t + \phi)}[/tex]

where [tex]\omega = \sqrt{2k/m}[/tex] and [itex]\phi[/itex] is a phase factor that depends on your choice of when t = 0.

What is the value of A?

AM
roman15
#7
Mar8-11, 10:20 AM
P: 70
wait, wouldnt w= squareroot(2ksin(theta)/m)?
because the sin(theta) i got for the acceleration just depends on the initial angle the mass starts from when displaced, then x = Asin(omegat + phi)
and im not sure what A would be
Andrew Mason
#8
Mar8-11, 11:38 AM
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Quote Quote by roman15 View Post
wait, wouldnt w= squareroot(2ksin(theta)/m)?
because the sin(theta) i got for the acceleration just depends on the initial angle the mass starts from when displaced, then x = Asin(omegat + phi)
No. If you take the second derivative of x with respect to time you get the acceleration:

[tex]\frac{d^2x}{dt^2} = A\frac{d}{dt}(\omega(\cos{\omega t + \phi})) = A\omega^2(-\sin{(\omega t + \phi})) = - \omega^2 x = a[/tex]

You have found that:

[tex]a = -2kx/m[/tex]

So that means that:

[tex]\omega^2 = 2k/m[/tex]

and im not sure what A would be
What is the maximum value of x?

AM
SammyS
#9
Mar8-11, 06:43 PM
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If the motion is perpendicular to the springs, the solution is quite a bit different.

In your figure this would be the case if the mass moved in a direction into and then out of the page -- away from the viewer then toward the viewer.
roman15
#10
Mar8-11, 10:23 PM
P: 70
oh so the equation for the acceleration that i got is for if the spring moves forwards and backwards, not up and down?
what do i do to make it for the mass going up and down?
SammyS
#11
Mar8-11, 11:19 PM
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If the mass moves a distance x from it's equilibrium position, then each spring stretches from 2L to √(4L2+x2).

The component of force in the x-direction is just x/√(4L2+x2) times the magnitude of the force.
Andrew Mason
#12
Mar9-11, 05:28 AM
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Quote Quote by roman15 View Post
from examining the forces I got that [itex]a(x)=(-2kx\sin{(\theta))}/m[/itex]
Please note: in your original answer the sin(theta) term is incorrect. I am not sure what [itex]\theta[/itex] is. As I have stated above, the correct equation for the acceleration of m is:

a(x)=-2kx/m

where x = [itex]\Delta x[/itex] - the displacement from the equilibrium position in the vertical direction.

AM
roman15
#13
Mar9-11, 02:42 PM
P: 70
hmm im going to ask my professor if we should assume the springs stay the same length when we displace the mass, because the question says were displacing it by small distances
and for the sin(theta), well if you displace the mass then the x component of the spring force would include sin(theta)
Andrew Mason
#14
Mar9-11, 03:52 PM
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Quote Quote by roman15 View Post
hmm im going to ask my professor if we should assume the springs stay the same length when we displace the mass, because the question says were displacing it by small distances
and for the sin(theta), well if you displace the mass then the x component of the spring force would include sin(theta)
Your question says that the displacement is in the vertical direction. If the springs are vertical, that means the springs compress and stretch vertically along the axes of the springs.

AM
roman15
#15
Mar9-11, 05:26 PM
P: 70
no the question doesnt say that the springs are vertical, and from the diagram i gave shows that the springs are horizontal, but the x direction is vertical
SammyS
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Mar9-11, 10:21 PM
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Quote Quote by SammyS View Post
If the mass moves a distance x from it's equilibrium position, then each spring stretches from 2L to √(4L2+x2).

The component of force in the x-direction is just x/√(4L2+x2) times the magnitude of the force.
The above quote goes with my previous post in which the mass moves perpendicular to the line along which the springs lie when the system is in equilibrium.


When the mass is displaced a distance x from equilibrium the magnitude of the force exerted by each spring is: [tex]F_s=k\left(\sqrt{4L^2+x^2}\,-\,L\right)[/tex]

Take the component of force in the x-direction & multiply by 2 to get the magnitude of the net Force. The net force, F, is in the negative x-direction.

[tex]F=2k\left(\sqrt{4L^2+x^2}\,-\,L\right)\frac{x}{\sqrt{4L^2+x^2}}=2kx\left(1-\frac{L}{\sqrt{4L^2+x^2}}\right)=2kx\left(1-\frac{1}{\sqrt{4+(x/L)^2}}\right)[/tex]

The power series expansion for [tex]\frac{1}{\sqrt{4+u^2}}[/tex] is [tex]\frac{1}{\sqrt{4+u^2}\,}= {{1}\over{2}}-{{x^2}\over{16}}+{{3 x^4}\over{256}}-{{5 x^6}\over{2048}}+\dots[/tex]

.
Andrew Mason
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Mar9-11, 11:25 PM
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Quote Quote by roman15 View Post
no the question doesnt say that the springs are vertical, and from the diagram i gave shows that the springs are horizontal, but the x direction is vertical
A simple statement saying the springs are stretched horizontally and the mass is displaced vertically would have made it clear. But I see you put that in the title. Missed that. In any event, this is not a problem because the solution is the same:

The restoring force on m is the vertical component of the spring force when the mass is displaced a distance x. The spring force is constant in magnitude (F = -k2L) but changes in direction. It is only the vertical component that you are concerned about. Let the angle that the springs make with the horizontal be [itex]\theta[/itex]. Since the springs are identical and have the same stretch and length, they provide the same force and make the same angle for a given displacement.:

[tex]F_x = -k(2L)sin\theta + -k(2L)sin\theta = -4kLsin\theta[/tex]

Since [itex]sin\theta = x/2L[/itex] the equation becomes:

[tex]F_x = -2kx[/tex]

So:

[tex]a = -2kx/m[/tex]

and, therefore the solution is the one I gave you above.

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SammyS
#18
Mar10-11, 07:35 AM
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My derivation come out to F = -2kx(1/2+...) ≈ -kx.

Check my derivation. I'll look later.


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